I am going to look at the barber version of Russell's paradox but same argument will work for sets version as well. The barber version goes as:

A barber cuts hair of only those people who do not cut hair of themselves. ---(1)

Paradox arises when one asks whether barber cuts his own hair or not. A further simplified version can be written as:

For all persons that exist, a barber cuts hair of a person if and only if that person does not cut hair of self. ---(2)

Now if we designate barber with B, a person with X and P cuts hair of Q as f(P,Q). Then (2) basically can be written as:

∀X, f(B,X) iff ¬f(X,X) ---(3)

So if we assume f(B,B) this will imply ¬f(B,B) and if we assume ¬f(B,B) this will imply f(B,B), which raises the paradox. Note that we need to assume either f(B,B) or ¬f(B,B), otherwise that will violate the Law of Excluded Middle. Now, note that (3) is a collection of all the following statements:

f(B,X1) iff ¬f(X1,X1)

f(B,X2) iff ¬f(X2,X2)

...

f(B,B) iff ¬f(B,B) ---(4)

...

f(B,Xn) iff ¬f(Xn,Xn)

The statement (4) is basically a violation of the Law of Non-Contradiction (not exactly, but one can prove it). Similarly, one can start from a logically invalid statement and reach Russell's Paradox:

f(B,B) iff ¬f(B,B) ---(5)

Now assume that all X apart from B satisfy:

∀X≠B, f(B,X) iff ¬f(X,X) ---(6)

Note that (6) is logically valid. So, we combine a logically invalid sentence (5) with logically valid sentence (6) to reach (3), which will then obviously be logically invalid too.

So the question is if Russell's Paradox is basically equivalent to the statement:

A iff ¬A ---(7)

so what is so great about it? In philosophical sense, what is the significance of Russell's Paradox?

If you state your original statement itself as (7) then assumption A will lead to ¬A and assumption ¬A will lead to A. Basically it does not seem like Russell's Paradox starts with something valid and logically concludes something invalid, rather it seems like it starts with something logically invalid in the first place, namely the violation of the basics of Logic.

There are numerous ways suggested to avoid Russell's Paradox by putting several kinds of restrictions; but following my argument above, we only need to disallow any statement that inherently contains (7), this way one can avoid Russell's Paradox in the set theory. Or am I misunderstanding something and the matter is more subtle than this?

Also, I will be very thankful if someone can tell me if similar discussion is done elsewhere in literature, so I can read more about it.

NOTE 1: If you think of function f(P,Q) as P contains Q then you get the sets version of Russell's Paradox. In fact, f(P,Q) can be any function for example if we choose it as "love" and B as Bess, then Russell's Paradox will become something like "Bess loves someone only if she doesn't love herself, does Bess love herself?"

In Barber language, "A barber cuts hair of only those people who don't cut the hair of barber, does barber cut his own hair?" And in Sets language, "A set P only contains those sets that don't contain P, does P contain P?" So again, the question is what is so special about Russell's Paradox and this one too if they are basically equivalent to (7)?

NOTE 3: The SEP Article initially points out issues with Unrestricted Comprehension Axiom and vicious circle but in my explanation above, I see problems with Russell's Paradox in its statement and assumptions itself. Later it discusses it in Contemporary Logic but none of that seems to be similar to my argument, unless I missed something.

NOTE 4: To give an analogy to the question: Assume we have a law:

1+1=2 ---(9)

then we make the following sentence:

Two and Two Apples are Six Apples ---(10)

Now, (10) reduces to 2+2=6, divided by 2 gives 1+1=3, which directly violates my law (9). Now what is so great about (10), do we need to restrict our comprehension to avoid making statements like (10)? Similarly we have some assumed laws of logic, if Russell's Paradox contains a statement (4) that directly violates the basic laws of logic then what is so great about it? It is not much different compared to making statements like (10).

NOTE 5: Because a lot of people are not satisfied with barber version I mentioned above, here I am adding the sets version of my argument. Russell's Paradox describes a set:

R = {x | x ∉ x} ---(11)

this can be written as:

∀ x, x ∈ R ⟺ x ∉ x ---(12)

equivalently:

∀ x, x ∈ R ⟺ ¬(x ∈ x) ---(13)

this is a collection of following sentences:

x1 ∈ R ⟺ ¬(x1 ∈ x1)

x2 ∈ R ⟺ ¬(x2 ∈ x2)

...

R ∈ R ⟺ ¬(R ∈ R) ---(14)

...

xn ∈ R ⟺ ¬(xn ∈ xn)

The statement (14) is same as what I meant above in (4). Rest of the question then follows as earlier.

The barber paradox is not Russell's paradox nor is it a version of Russell's paradox.
– user4894Jan 17 '18 at 0:16

1

@user4894, it is said so in various places like Scientific American. The first line of Wikipedia page on Barber Paradox gives a reference. In my whole argument you can put f(P,Q) as P contains Q and you get the usual sets version of Russell's Paradox.
– udy11Jan 17 '18 at 5:27

It has a family resemblance to Russell's paradox...
– Mozibur UllahJan 17 '18 at 5:54

And yes: the Paradox assume Non Contradiciton and Bivalence: "Since by classical logic one case or the other must hold – either R is a member of itself or it is not – it follows that the theory implies a contradiction."
– Mauro ALLEGRANZAJan 17 '18 at 7:06

4 Answers
4

Not sure about the question... but we may prove "by logic alone" that:

¬(∃y)[(∀x)(A(x,y) ↔ ¬A(x,x))].

Proof:

1) (∃y)[(∀x)(A(x,y) ↔ ¬A(x,x))] --- assumed

2) (∀x)(A(x,c) ↔ ¬A(x,x)) --- where c is a new constant

3) A(c,c) ↔ ¬A(c,c) --- by universal instantiation.

The formula A(c, c) ↔ ¬A(c, c) is, or implies, a contradiction, since obviously A(c, c) ↔ A(c, c), and thus we conclude with the negation of the assumption.

Thus, if your question is about a "pattern" common to Barber's paradox as well as Russell's antinomy, the answer is YES. See Cantor' diagonal argument.

The common "pattern" shows that a certain existential assumption is untenable, due to the fact that it leads to a contradiciton.

Thus, for the Barber version, the conclusion is: person B does not exists.

In the same way, for the set version: the set R does not exist.

But note that the two above conclusions are not "paradoxical" at all; they are perfect sound examples of proofs by contradiction. They disprove an initial conjecture: the existence of barber B (of set R) deriving a contradiction.

Where is the difference ? In the role of the Comprehension Axiom, and it is a big one.

In the Barber case, we stay happy with the proof that the purported barber does not exists: this conclusion per se has no mathematical or philosophical impact.

Regarding the original set theory, instead, the impact is relevant.

The unrestricted versione of the Comprehension Axiom was very natural and uncontroversial: for every property P there is a collection of things that have that property: { x ∣ P(x) }.

The use of the axiom in the definition of the Russell's set R shows that the natural and uncontroversial principle is not teneable: end of the story.

thanks for SEP article link. i have also expanded my original post to clarify the query and include more examples. i am also basically saying the same argument that you have given here, and so this raises the question that if Russell's Paradox is equivalent to A iff ¬A then what is so special about it?
– udy11Jan 17 '18 at 20:04

@udy11 Any paradox is equivalent to A <-> ~A; that's just what it means to be paradoxical. The question is, what is wrong with this concept such that it is a paradox?
– CanyonJan 17 '18 at 20:07

@Canyon yes a paradox is when you logically conclude something invalid from logically valid assumptions but Russell's Paradox seems like stating an invalid statement in the first place (or so you can prove in 2 lines as I did in original post), so what is so special if you state something illogical and derive illogical thing from it and call it a paradox? furthermore, if it's not that significant then why so many attempts at mending set theory have been made, one could have simply disallowed all statements that contain or reduce to A <-> ~A?
– udy11Jan 17 '18 at 20:20

* sorry if it sounds wrong, but i know i am unaware of many things or mistaken somewhere and that is why just asking questions. please don't mind my poor choice of words
– udy11Jan 17 '18 at 20:27

@udy11 - The problems with set theory (it seems to me) do not come from the barber paradox but from problems of self-reference raised by the set-of-all-sets. This is much more like a paradox than the barber (or the liar) example. It seems to me that the barber situation is not equivalent to Russell's paradox. But it's not unconnected so you might have a point. There no paradox if we dismiss the set-of-all-sets as an impossible object (which is my preferred approach). If we say it is a possible object than metaphysics runs into terminal problems.
– PeterJJan 22 '18 at 10:38

Today, Russell's paradox is simply the proof in ZFC the class R = {x ∣ x ∉ x} is not a set. One avoids the paradox because you cannot meaningfully write f(R,R).

But in Russell's time, people were using the axiom of unrestricted comprehension, and together with the methodology of the day, this implies that R really is a set, and thus f(R,R) is meaningful. But since even being able to write f(R,R) leads to a contradiction, this implies there is something seriously wrong with the set theory of the time.

It's maybe worth noting that Russell's paradox has a direct analog in naive formal logic as well. If predicates can take arbitrary predicates as arguments, then if X is a unary predicate it makes sense to feed it to itself. We can define a predicate P by

P(X) := ¬X(X)

and then derive a contradiction by contemplating P(P).

Just as set theory switched over to adopted restricted comprehension, formal logic had to adapt as well, so that it no longer implies that you can define such a P.

For example, the typical approach is to stratify logic into first-order logic which can reason about objects, second-order logic which can reason about first-order predicates, and so forth. In this way, it never makes sense to write X(X), since the argument to any unary predicate X must have lower order than X itself.

agreed with this. i have expanded my original post to clarify my point, so the question is somewhat different
– udy11Jan 17 '18 at 20:47

1

@udy11: I don't think you've changed anything. You still seem to be missing the point that Russell's paradox was "The formalization of set theory says that this set exists. But this set existing leads to contradiction". Anyways, I've added to my answer an analog of Russell's paradox in formal logic.
– HurkylJan 17 '18 at 21:42

"that" set exists is violation of basic laws of logic, namely A iff ¬A, so you mean to say we needed to avoid this contradiction that we adopted restricted comprehension. my question is why not just say that anything that violates basic logic is not allowed?
– udy11Jan 18 '18 at 7:08

"Basically it does not seem like Russell's Paradox starts with
something valid and logically concludes something invalid, rather it
seems like it starts with something logically invalid in the first
place, namely the violation of the basics of Logic."

It is not just that "it seems", it is indeed the case that Russell's Paradox starts with something logically invalid in the first place, namely a logically inconsistent universe, as seen after (3). An alternative way to perceive it is by noting that Russell's Paradox is plainly incompatible with the axioms of a logically self-consistent universe.

A logically self-consistent one-tidy-and-respectful-barber universe is based on these two axioms:

The barber cuts his hair himself.

For x = B, f(B,x).

The barber cuts the hair of any other person iff that person does not cut his hair himself.

For all x ¬= B, f(B,x) XOR f(x,x).

Since any boolean variable can be expressed as either an AND or an OR with itself, and since in axiom 1 f(B,x) = f(x,x), it can be expressed as either:

1.a. For x = B, f(B,x) AND f(x,x).

1.b. For x = B, f(B,x) OR f(x,x).

From axioms 1.a and 2, or 1.b and 2, it is clear that no valid proposition can be stated for all x in a logically self-consistent one-tidy-and-respectful-barber universe. So Russell's Paradox cannot happen.

In the example of "Bess loves someone only if she doesn't love herself", a logically self-consistent Bess-loving-herself-and-the-not-self-loving universe is based on these two axioms:

i can easily modify the statement. my main point was to deliver the idea. for my question, you can then instead consider the sets version or Bess-love version I mentioned in NOTE 1
– udy11Jan 20 '18 at 20:46

Why not just say that anything that violates basic logic is not allowed?

There is indeed a rule saying that if a set of statements generates a contradiction, the conjunction of those statements is false. We do not allow anything that violates basic logic.

How can we have Russell's Paradox, then? The problem is, we have no notion of the priority of rules inside a particular formal language. So while our principle of non-contradiction tells us that

NC: P_1, ..., P_n |- _|_ => ~(P_1 & ... & P_n)

we also have (in naive set theory)

A_1, ..., A_n

such that

A_1, ..., A_n |- _|_

which is clearly bad, as well as

NC, A_1, ... A_n |- _|_

which is even worse.

Unfortunately truth is defined inside a particular language. Therefore we cannot tell within the language itself which one is in the wrong: NC or any one of A_1 to A_n.

Instead of throwing out particular theorems because they are paradoxical, we throw out sets of rules that generate those theorems. The new set of rules in the case of set theory had the axiom of restricted comprehension which avoids the problem.